The Mean Value Theorem and Rolle's theorem
Problem 3.765 · medium
Verify that \( \displaystyle f(x) = x^{3} + x - 1 \) satisfies the hypotheses of the Mean Value Theorem on \( \displaystyle [0, 4] \), and find every number \( \displaystyle c \) the theorem guarantees.
- f is a polynomial, so it is continuous on [a, b] and differentiable on (a, b): the hypotheses hold.
- \[ \frac{\left(-1\right) \left. x^{3} + x - 1 \right|_{\substack{ x=0 }}}{4} + \frac{\left. x^{3} + x - 1 \right|_{\substack{ x=4 }}}{4} = 17 \]The slope of the secant line.✓ Proved
- \[ \frac{d}{d x} \left(x^{3} + x - 1\right) = 3 x^{2} + 1 \]Differentiate.✓ Proved
- \[ \left. 3 x^{2} + 1 \right|_{\substack{ x=\frac{4 \sqrt{3}}{3} }} = 17 \]c = 4*sqrt(3)/3 solves f′(c) = 17 and lies in (0, 4).✓ Proved
Answer \( c = \frac{4 \sqrt{3}}{3} \)
Lines: 3 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt. How we verify
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Not checked | — | a sentence; read, not computed |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | each c checked by a difference quotient; a scan of (a, b) finds no others |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (error) — The solution fails to find 'every' number c guaranteed by the theorem. The equation 3c^2 + 1 = 17 has two solutions, c = ±4√3/3, but only the positive one is listed. The negative solution is extraneous to the interval, but the solution does not explicitly state that the negative root is discarded because it lies outside (0, 4), nor does it acknowledge the existence of the second root before discarding it, which is incomplete reasoning for 'finding every number'.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-11-01 — The solution fails to find 'every' number c guaranteed by the theorem. The equation 3c^2 + 1 = 17 has two solutions, c = ±4√3/3, but only the positive one is listed. The negative solution is extraneous to the interval, but the solution does not explicitly state that the negative root is discarded because it lies outside (0, 4), nor does it acknowledge the existence of the second root before discarding it, which is incomplete reasoning for 'finding every number'.gpt-oss:20b: pass 2026-11-01qwen3.6:27b-mlx: inconclusive 2026-10-31 — reviewer returned a non-objectgpt-oss:20b: pass 2026-10-31qwen3.6:27b-mlx: fail (error) 2026-10-31 — The solution fails to find all numbers c guaranteed by the theorem. Solving 3c^2 + 1 = 17 yields c = ±4√3/3, but only the positive root is listed. The negative root is not in the interval, but the solution does not explicitly state that the negative root is rejected because it lies outside (0, 4), nor does it acknowledge the existence of the second algebraic solution before discarding it.gpt-oss:20b: pass 2026-10-31
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
generator:structured/mean_value_theorem, checked 2026-11-01 with SymPy 1.14.0.